Introduction
A trigonometric expression such as \(\sin(1/x)\) oscillates faster and faster as \(x\) approaches zero from the right. It does not settle toward a single value. Yet multiplying it by a positive power of \(x\) can force the entire product toward zero.
This remains true when the oscillating factor includes a fixed offset, as in \(c+\sin(1/x)\) or \(c+\cos(1/x)\). The offset can make the factor stay positive, stay negative, or change sign. What matters is that the factor remains bounded while \(x^p\) approaches zero.
The Bounds for an Offset Oscillation
For every real number \(t\), the trigonometric functions satisfy \(-1\le \sin t\le 1\) and \(-1\le \cos t\le 1\). Adding a fixed real constant \(c\) gives
and
These inequalities still hold when \(t=1/x\). For \(x>0\), the multiplier \(x^p\) is positive, so multiplying each inequality by \(x^p\) preserves its order. If \(p>0\), then \(x^p\to0\) as \(x\to0^+\). The lower and upper bounds therefore both approach zero.
The offset \(c\) can be positive, zero, or negative. The theorem does not require the expression in parentheses to have one sign. It only uses the fact that it stays between two fixed numbers.
Why the Squeeze Theorem Applies
The Squeeze Theorem states that if \(g(x)\le f(x)\le h(x)\) for all \(x\) sufficiently close to a limit point (with \(x\ne0\)), and if \(g(x)\) and \(h(x)\) both approach the same limit \(L\), then \(f(x)\) also approaches \(L\).
For the sine expression, choose \(g(x)=x^p(c-1)\), \(f(x)=x^p(c+\sin(1/x))\), and \(h(x)=x^p(c+1)\). The trigonometric bounds give \(g(x)\le f(x)\le h(x)\) whenever \(x>0\). Since \(c-1\) and \(c+1\) are fixed constants and \(p>0\),
The two bounds have the same limit, so the Squeeze Theorem gives the claimed limit of \(f(x)\). For the cosine expression, the inequality \(-1\le\cos(1/x)\le1\) gives the same lower and upper bounds, so the same argument proves its limit.
Use \(-1\le\sin(1/x)\le1\), or \(-1\le\cos(1/x)\le1\), and include the fixed offset.
Because \(x\to0^+\), we have \(x>0\), so \(x^p>0\) for every real \(p>0\).
Each bound is a fixed constant times \(x^p\), so each approaches zero.
The expression between the bounds must approach their common limit, zero.
Worked Examples
Worked Example: A Positive Offset That Keeps the Factor Positive
Evaluate \(\displaystyle\lim_{x\to0^+}\sqrt{x}\bigl(4+\sin(1/x)\bigr)\).
Since \(-1\le\sin(1/x)\le1\), adding \(4\) gives
For \(x>0\), \(\sqrt{x}>0\), so multiplying through by \(\sqrt{x}\) preserves the order:
The lower bound has limit \(3\cdot0=0\), and the upper bound has limit \(5\cdot0=0\), because \(\sqrt{x}\to0\). The Squeeze Theorem gives
Check: The expression is between \(3\sqrt{x}\) and \(5\sqrt{x}\), and both bounds approach zero. Thus the result is consistent with the inequalities.
Worked Example: A Factor That Can Be Zero
Evaluate \(\displaystyle\lim_{x\to0^+}x^2\bigl(1+\cos(1/x)\bigr)\).
The cosine bounds give \(-1\le\cos(1/x)\le1\). Adding \(1\) yields
Multiplying by \(x^2>0\) gives
The left-hand bound is constantly zero, so its limit is zero. The right-hand bound has limit \(2\cdot0=0\). Therefore, by the Squeeze Theorem,
Check: The expression is nonnegative and no larger than \(2x^2\). Since \(2x^2\to0\), the limit zero agrees with both bounds.
Worked Example: A Negative Offset
Evaluate \(\displaystyle\lim_{x\to0^+}x^{3/2}\bigl(-2+\cos(1/x)\bigr)\).
Starting from \(-1\le\cos(1/x)\le1\), subtracting \(2\) gives
The multiplier \(x^{3/2}\) is positive for \(x>0\), so the order remains unchanged:
The left bound approaches \(-3\cdot0=0\), and the right bound approaches \(-1\cdot0=0\). Thus the Squeeze Theorem gives
Check: The expression is negative or zero near the limit point, but it is trapped between two quantities that both approach zero. A negative expression can therefore have limit zero.
Why the Oscillation Does Not Need Its Own Limit
It is tempting to try to find a limit for the trigonometric factor first. In general, that limit does not exist. For instance, consider the two positive sequences
where \(n\) is a positive integer. Both sequences approach zero. But \(\sin(1/x_n)=1\), whereas \(\sin(1/y_n)=-1\). Thus \(\sin(1/x)\) does not approach a single value as \(x\to0^+\). Nevertheless, for any \(p>0\) and fixed \(c\), the product \(x^p(c+\sin(1/x))\) approaches zero, as the bounds prove.
This distinction is important: a factor can keep oscillating without a limit, while a multiplier that approaches zero controls the size of the product. The Squeeze Theorem establishes the product’s limit without requiring a limit for the oscillating factor.
A Common Mistake: Treating Infinity as an Input
A common incorrect calculation is to write
This is not valid. Infinity is not a real input at which sine can be evaluated, and the sine values do not approach zero: the sequences above give values \(1\) and \(-1\). Therefore, it is also not valid to claim the product’s limit by multiplying a supposed limit of the sine factor by another limit.
The correction is to bound the whole offset factor instead:
After multiplying by \(x^p>0\), the bounds become \(x^p(c-1)\) and \(x^p(c+1)\), both of which approach zero. The Squeeze Theorem then proves the product limit without assigning a limit to \(\sin(1/x)\).
Conditions and Scope
The exponent condition \(p>0\) matters because it ensures \(x^p\to0\) as \(x\to0^+\). If \(p=0\), the factor \(x^p\) is \(1\), so the oscillation is not damped; for example, \(1+\sin(1/x)\) has no limit as \(x\to0^+\). If \(p<0\), then \(x^p\) grows without bound as \(x\to0^+\), so the argument using two bounds that approach zero no longer applies.
The right-hand limit is used because \(x^p\) is defined as a real-valued power for \(x>0\) for every real \(p\). When \(p\) is a positive integer, the corresponding two-sided limit is also zero: the absolute value of the product is at most \((|c|+1)|x|^p\), which approaches zero from either side. The stated result for arbitrary positive real \(p\) requires only the right-hand limit.
Check Your Understanding
Use bounds and the Squeeze Theorem where appropriate. State why the bounds apply.
- Evaluate \(\displaystyle\lim_{x\to0^+}x^{1/3}\bigl(2+\sin(1/x)\bigr)\).
- Evaluate \(\displaystyle\lim_{x\to0^+}x^4\bigl(-3+\cos(1/x)\bigr)\).
- For a fixed real \(c\) and \(p>0\), give lower and upper bounds for \(x^p(c+\sin(1/x))\) when \(x>0\).
- Why does the proof not require \(\sin(1/x)\) to have a limit?
- What changes if the exponent is \(p=0\) instead of \(p>0\)?
